(New page: This problem is not very hard to prove. Especially if you read then end of page 387 and the beginning of page 388 (The proof is basically there). First off: Suppose F is a subfield of GF(...) |
|||
| Line 4: | Line 4: | ||
n = [GF(p^n) : GF(p)] | n = [GF(p^n) : GF(p)] | ||
| + | |||
n = [GF(p^n) : GF(p^m)]*[GF(p^m) : GF(p)] | n = [GF(p^n) : GF(p^m)]*[GF(p^m) : GF(p)] | ||
| + | |||
n = [GF(p^n) : GF(p^m)]*m | n = [GF(p^n) : GF(p^m)]*m | ||
Latest revision as of 03:51, 30 November 2012
This problem is not very hard to prove. Especially if you read then end of page 387 and the beginning of page 388 (The proof is basically there).
First off: Suppose F is a subfield of GF(p^n) then F is isomorphic to GF(p^m) for some arbitrary m. By theorem 21.5 it follows that since m divides n then
n = [GF(p^n) : GF(p)]
n = [GF(p^n) : GF(p^m)]*[GF(p^m) : GF(p)]
n = [GF(p^n) : GF(p^m)]*m
And thus [GF(p^n) : GF(p^m)]= n/m
--Bakey 08:50, 30 November 2012 (UTC)
